AIC with ALSs

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AIC with ALSs
A Locked Set is a group of cells (that can all see each other) of size N where the number of candidates in those cells is equal to the size of the group. That is N cells contain N candidates. A solved cell or a clue is a Locked Set where N=1, but such a cell is not useful. The smallest useful Locked Set is a Naked Pair (where N=2) followed by a Naked Triple (N=3).

An Almost Locked Set (ALS) is N cells containing N+1 candidates. In the context of Alternating Inference Chains in this solver, an ALS is of size N=2 and the number of different candidates in those cells is 3, although bigger ALS groups are possible. So an ALS of size 2 will be a two Conjugate Pairs plus one other candidate.

This might be easier to explain with a diagram. The highlighted yellow cells D7 and D8 contain [2,7] and [2,7,9] respectively. If D8 didn't contain a 9 the two cells would be a Naked Pair. The extra 9 makes these two cells an ALS.

While solving a puzzle I am hunting around for Inference Chains and perhaps I start with the bi-value cell J2 containing [2,9]. I pretend to turn ON the 2 which turns OFF the 9 in that cell. The only other 9 is in J8 which I turn ON. Looking up the column the 9 in D8 must be turned OFF. This creates an on-the-fly Naked Pair.

Now, a Naked Pair eliminates candidates in the row or column (or box) it is aligned on so we can use this elimination property as part of our chain. This is the trick! By removing the 9 in D8 we fix 2 and 7 into those two cells so we can look along the row at other 2s and 7 and turn them OFF. This I do in cell D3. From there I can continue the inference chain.

ALS in an AIC fragment

A real life example now. This chain contains an ALS on the cells {G6,H6} (I used squiggly brackets to denote ALS as opposed to square brackets for Grouped Cells). 9s in row H are the entry point. We turn 9 ON in H2 which turns OFF the 9 in H6 - the extra candidate that makes the ALS an ALS. This gives us a Naked Pair of [5,7] that points up column 6 turning OFF the 7 in F6 and the chain continues.

Ultimately we use Nice Loop Rule 2 to place 4 in A4
AIC on 4 (Discontinuous Alternating Nice Loop, length 12):
-4[A4]+4[D4]-7[D4]+7[D2]
-7[H2]+9[H2]-9[H6]+7{H6|G6}
-7[F6]+4[F6]-4[A6]+4[A4]
- Contradiction: When 4 is removed from A4 the chain implies it must be 4 - other candidates 2/5 can be removed

AIC with ALS: Load Example or : From the Start

This second example uses a chain to kill off-chain candidates, which is Nice Loop Rule 1. The ALS is in {F1,F4} and consists of [1/3/8] and [1/3] respectively. We turn off the extra candidate, 8 in F1 to enable the Naked Pair to be formed.

AIC (Alternating Inference Chain) Rule 1:
+3[B3]-3[B5]+6[B5]-6[B8]+6[D8]
-8[D8]+8[D1]-8[F1]+3{F1|F4}-3[F3]+3[B3]
- Off-chain candidate 3 taken off B4 - weak link: B3 to B5
- Off-chain candidates 1 taken off cell D8, link is between 6 and 8 in D8
- Off-chain candidate 8 taken off F2 - weak link: D1 to F1
- Off-chain candidate 8 taken off F3 - weak link: D1 to F1

Off Chain eliminations: Load Example or : From the Start

What's interesting about this Inference Chain is the elimination of 8s in F2 and F3 despite the chain itself containing an OFF 8 in F1. Basically it looks like there are no 8s left in row F but of course the Inference Chain doesn't tell us the ultimate fate of any of the elements that make it, merely the properties of other candidates that the chain as a whole proposes. But at a glance we know F1 will be 8 because it is the last 8 in the row.

Go back to AICs with Grouped Cells

Continue to Digit Forcing Chains

Comments

Comments - Talk

Wednesday 3-Oct-2012

... by: Str8tsFan

I hope Andrew will ever read (and answer) these comments... About the second example:

(1) Using the link to the solver leads to a sudoku with a tiny little difference: B9 has an additional candidate 6, which is missing at the example. At the solver this candidate will be eliminated with the very same example:
"- Off-chain candidate 6 taken off B9 - weak link: B5 to B8"

(2) What about the candidate 8 at J1? As far as I can understand the theory, the weak link "D1 to F1" should not only eliminate the 8s at F2 and F3 (weak link in same box) but also the 8 at J1 (weak link in same column). More interesting: the solver doesn't eliminate that 8 at J1 either. Why? Did I make any mistake, or is it a flaw of the solver? As far as I understood, a weak link can be part of two entities (box and row or box and column) and thus should be able to eliminate candidates at both entities, maybe the solver fails to check the second entity?

Tuesday 10-Jul-2012

... by: Mr Turner

The last paragraph seems flawed since there is an 8 at F8. The 8s at F2 and F3 could be colored off as an inference from D1 being on. This implies F8 is on. F8 is also inferred to be on from D8 being off.

Wednesday 22-Apr-2009

... by: Mr Archibald

I call this the 6 pack.

the first three in a line on rows 1,2 is the some as the last six on row 3. this can work upsidedown and backtofront and sideways.

a b c d e f g h i
i h g f e d c b a
d e f a b c i h g